# What is the probability that the first ace in a deck is the 30th card?

For each potential card there is a $\frac{12}{13}$ chance that it isn't an ace. Therefore in order for the first ace to occur at position 30, the first 29 cards can't be an ace. The probability of this is $(\frac{12}{13})^{29}$.

The probability of card 30 being an ace among the 23 remaining cards in the deck is $\frac{4}{23}$, and so the probability of card 30 being the first ace is $(\frac{12}{13})^{29} * \frac{4}{23} \approx 0.017$.

Is my reasoning correct?

• not really. observe that if you already know that the first card isn't an ace then the proportions in the remaining cards change, you only have 51 instead of 52 cards to choose from – mm-aops Oct 6 '14 at 19:50

## 4 Answers

The positions of the four aces is a random subset of size four taken from the index set $1,2,\dots, 52$. There are ${52\choose 4}$ equally likely outcomes.

The number of outcomes that include the index "30" and three larger indices is $22\choose 3$, since there are 22 indices in the range $31,32,\dots, 52$.

The required probability is $${{22\choose 3}\over{52\choose 4}}={44\over 7735}=0.005689.$$

The probability that the second card isn't an ace is not $\frac{12}{13}$. Since we know the first card was not an ace, there are 4 aces and 47 non-aces. Hence, the probability that the second card is not an ace is $\frac{47}{51}$.

$$\frac{48}{52}\cdot\frac{47}{51}\cdot\frac{46}{50}\cdots\frac{20}{24}\cdot\frac{4}{23}=\frac{22\cdot21\cdot20\cdot4}{52\cdot51\cdot50\cdot49}\approx0.57\%$$

• What about "30"? Is there a flow in my reasoning? – Yulia V Oct 6 '14 at 19:57
• @YuliaV See comment to your answer. (Note: "flaw", not "flow".) – Did Oct 6 '14 at 20:05
• @Did Ahh I understand my mistake. What if I have to find the original probability given that the 31st card is an ace? Could it be $\frac{48}{51} * \frac{47}{50} ... * \frac{20}{23} * \frac{3}{22} \approx 1\%$ – Shen Nung Oct 6 '14 at 21:03
• The probability that the 30th card is the first ace given that the 31st card is an ace? – Did Oct 6 '14 at 21:05
• Yes. I was wondering if I could just slightly change the formula you provided or if I'd need to use something like Bayes' formula. – Shen Nung Oct 6 '14 at 21:09

Another visualization, leading (I hope) to the correct result already given.

Consider a row of $52$ empty, ordered slots. You're going to randomly selects a slot for each of the $52$ cards, and you may as well put the four aces in first.

To have the $30$th slot be the first ace, we first need to have the first $29$ slots ace-free. The first ace has $52$ slots to go into, of which $23$ are not in the first $29$, so it will be in the correct zone with probability $\frac{23}{52}$. Assuming that this first ace misses the forbidden zone, the second ace has $51$ slots, of which $22$ are not in the first $29$, and so on. After placing all four aces, the probability of an ace-free first $29$ is $P_1$:$$P_1=\frac{23\times22\times21\times20}{52\times51\times50\times49}$$

Assume that the first $29$ slots are indeed ace-free. What is the probability that all four aces have missed Slot #$30$? The first ace has, we know, landed in one of $23$ slots; $22$ of them are not Slot #$30$, so this first ace misses Slot #$30$ with probability $\frac{22}{23}$. The second ace has only $22$ slots, of which $21$ are not Slot #$30$ and so on. The probability that all four aces miss Slot #$30$ is $P_2$:$$P_2=\frac{22\times21\times20\times19}{23\times22\times21\times20}=\frac{19}{23}$$So the probability that one of the four aces will hit Slot #$30$ is $1-P_2$, or $\frac{4}{23}$ So the overall probability of finding the first ace in Slot #$30$ is $$P=P_1 \times \frac4{23}=\frac{23\times22\times21\times20\times4}{52\times51\times50\times49\times23}$$